# Interesting problem on "neighbor fractions"

This is from I. M. Gelfand's Algebra book.

Fractions $$\displaystyle\frac{a}{b}$$ and $$\displaystyle\frac{c}{d}$$ are called neighbor fractions if their difference $$\displaystyle\frac{ad - bc}{bd}$$ has numerator of $$\pm 1$$, that is, $$ad - bc = \pm 1$$. Prove that

(a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);

(b) if $$\displaystyle\frac{a}{b}$$ and $$\displaystyle\frac{c}{d}$$ are neighbor fractions, then $$\displaystyle\frac{a+c}{b+d}$$ is between them and is a neighbor fraction for both $$\displaystyle\frac{a}{b}$$ and $$\displaystyle\frac{c}{d}$$; moreover,

(c) no fraction $$\displaystyle\frac{e}{f}$$ with positive integer $$e$$ and $$f$$ such that $$f < b + d$$ is between $$\displaystyle\frac{a}{b}$$ and $$\displaystyle\frac{c}{d}$$.

Parts (a) and (b) weren't too difficult, but I'm stuck on part (c). I've included (a) and (b) in case they're related to the solution to (c).

• Look up "Farey sequence". Jun 5, 2011 at 19:05
• Okay, I'll do that. Thanks! Jun 5, 2011 at 19:09

Assume $\frac{e}{f}$ is (strictly) between $\frac{a}{b}$ and $\frac{c}{d}$. Then $\left|\frac{a}{b}-\frac{e}{f}\right| + \left|\frac{e}{f}-\frac{c}{d}\right| = \left|\frac{a}{b}-\frac{c}{d}\right| = \frac{1}{bd}$
But $\left|\frac{a}{b}-\frac{e}{f}\right| \geq \frac{1}{bf}$ and $\left|\frac{e}{f}-\frac{c}{d}\right|\geq \frac{1}{df}$. So $\frac{1}{bf} + \frac{1}{df} \leq \frac{1}{bd}$. Multiply both sides by $bdf$ and we get that $b+d\leq f$.
• There is a geometric meaning to this, which is related to lattice points on the plane. If $(a,b)$ and $(c,d)$ are points on the plane, then $|ad-bc|$ is the area of the quadrilateral with points $(0,0)$, $(a,b)$, $(c,d)$ and $(a+c,b+d)$. If $a,b,c,d$ are integers, and $|ad-bc| = 1$, this theorem essentially says that there is no lattice point in the inside of the quadrilateral. Jun 5, 2011 at 20:26
• In this case, I wrote "strictly' precisely because I wanted to be clear that I meant $\frac{a}{b}<\frac{e}{f}<\frac{c}{d}$ or $\frac{c}{d}<\frac{e}{f}<\frac{a}{b}$. @ArshJhaj The theorem is not true if you allow $e=c,f=d$. The first part of my line is true for $\frac{e}{f}=\frac{a}{b}$, but the next step: $$\left|\frac{a}{b}-\frac{e}{f}\right|\geq \frac{1}{bf}$$ is obviously not Jul 8, 2015 at 10:59